Problem: $g(x)=4x^2+3$ What is the average rate of change of $g$ over the interval $[1,t]$, in terms of $t$, where $t\neq 1$ ? Your answer must be fully expanded and simplified.
Answer: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $g(1)=7$. We are interested in the average rate of change of $g(x)=4x^2+3$ over the interval $[1,t]$ : $\begin{aligned} &\phantom{=}\dfrac{g(t)-g(1)}{(t)-(1)} \\\\ &=\dfrac{4t^2+3-(7)}{t-(1)} \\\\ &=\dfrac{4t^2-4}{t-1} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{4t^2-4}{t-1}&=\dfrac{4(t^2-1)}{t-1} \\\\ &=\dfrac{4(t-1)(t+1)}{t-1} \\\\ &=4(t+1)\text{, for }t\neq 1 \\\\ &=4t+4\text{, for }t\neq 1 \end{aligned}$ Since we are given that $t\neq 1$, the average rate of change of the function is $4t+4$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.